What is the integral of (x^{1.3}+9x^{3.5}) ?

The integral of (x^{1.3}+9x^{3.5}) is 0.43478…x^{2.3}+2x^{4.5}+C

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integral of (x^{1.3}+9x^{3.5})

\int (x^{1.3} + 9 x^{3.5}) d x

0.43478 \dots x^{2.3} + 2 x^{4.5} + C

Solution steps

\int x^{1.3} + 9 x^{3.5} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x^{1.3} d x + \int 9 x^{3.5} d x

\int x^{1.3} d x = 0.43478 \dots x^{2.3}

\int 9 x^{3.5} d x = 2 x^{4.5}

= 0.43478 \dots x^{2.3} + 2 x^{4.5}

Add a constant to the solution

= 0.43478 \dots x^{2.3} + 2 x^{4.5} + C

\int \frac{5}{( x - 3 ) ( x + 2 )} d x

\frac{d}{d x} (\frac{4 x}{1 - cot ( x )})

\frac{2}{4 - x}

\frac{d}{d x} (x - π)

y = \frac{1}{x ^{4}}, (- 2, \frac{1}{16})

What is the integral of (x^{1.3}+9x^{3.5}) ?
The integral of (x^{1.3}+9x^{3.5}) is 0.43478…x^{2.3}+2x^{4.5}+C